Question

A single die is rolled. Find the probability of rolling an oddnumber or a number less than 6.

Answer 1

The probability of rolling an odd number or a number that is less than 6 is 5/12

Probability is the likelihood of the occurrence of an event and is the ratio of the number of favorable outcomes to the total number of outcomes of an event.

The sample space or possible outcomes when rolling a die is S = {1, 2, 3, 4, 5, 6}

We can say, n(S) = 6

Consider X as an event of getting an odd number

X = {1, 3, 5}

n (X) = 3

So P (X) = n(X)/ n(S)

By substituting the values,

P(X) = 3/6 = 1/2

Consider Y as an event of getting a number less than 6

Y = {1, 2, 3, 4, 5}

n (Y) = 5

So P (Y) = n(Y)/ n(S)

Substituting the values

P(Y) = 5/6

The probability of rolling an odd number or a number less than 6 = 1/2 × 5/6 = 5/12

Hence, the probability of rolling an odd number or a number less than 6 is 5/12.

Learn more about the probability of dice:

brainly.com/question/10298360

Answer 2

Given

A single die is rolled.

To find the probability of rolling an odd number or a number less than 6.

Explanation:

It is given that,

A single die is rolled.

Then, the sample space is,

$\begin{gathered} S=\lbrace1,2,3,4,5,6\rbrace \\ n(S)=6 \end{gathered}$

Let A be the event of getting an odd number.

Then,

$\begin{gathered} A=\lbrace1,3,5\rbrace \\ n(A)=3 \end{gathered}$

Let B be the event of getting a number less than 6.

Then,

$\begin{gathered} B=\lbrace1,2,3,4,5\rbrace \\ n(B)=5 \end{gathered}$

Also,

$\begin{gathered} A\cap B=\lbrace1,3,5\rbrace \\ n(A\cap B)=3 \end{gathered}$

Therefore,

$\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)} \\ =\frac{3}{6}+\frac{5}{6}-\frac{3}{6} \\ =\frac{3+5-3}{6} \\ =\frac{5}{6} \end{gathered}$

Hence, the answer is 5/6.